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I have a Quantum SDE containing both white and color noises (open quantum system). $$ \dot\rho(t) = A\rho_s + (\nu_{1t}\hat{c}^\dagger \hat{X}^-_1 + \omega_{1t} \hat{X}^+_1 \hat{c})\rho_s +(\nu_{2t} \hat{X}^+_2\hat{c} -\omega_{2t}\hat{c}^\dagger \hat{X}^-_2)\rho_s $$ Here A is a constant, $\hat{c}^\dagger$ and $\hat{c}$ are creation and annihilation operators belong to system. $X^\pm_j$ are operators belong to fermionic environment. The order of $X^\pm_j$, $\hat{c}^\dagger$ and $\hat{c}$ is fixed (no commutation or anti-commutation relation exist between $X^\pm_j$ and $\hat{c}^\dagger$, $\hat{c}$ ). $\nu_j$ is white noise, $\omega_{1t}$ and $\omega_{2t}$ are defined as $$ \omega_1(t) = \int_{t_0}^{t} C^+(t-\tau)\nu_{1\tau} d\tau$$ and $$ \omega_2(t) = \int_{t_0}^{t} C^-(t-\tau)\nu_{2\tau} d\tau $$ Where $C^\pm$ is defined as
$$C^\sigma(t)=\int_{-\infty}^{\infty} d\omega e^{i\sigma\omega t} f(\omega) J(\omega)$$ My question here is that how can we decrease the number of noises by combining white noises $\nu_j$ and color noises $\omega_j$ while they are connected to different operators. I know how to decrease the number of noises if they are connected to same operators. Any idea about different operator case, any article, book ? Your help will be appreciated.

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  • $\begingroup$ Can you give an example how you "decrease the number of noises" for the same operator case? It might help to clarify your question. $\endgroup$
    – user619894
    Dec 26, 2018 at 13:26
  • $\begingroup$ For example, For a case where $\nu_{1t}$ and $\omega_{1t}$ are connected to same operators, $\nu_{1t}$ and $\omega_{1t}$ can be replaced by one noise, i.e , $$\dot\rho(t) = A\rho_s + (\nu_{1t}\hat{c}^\dagger \hat{X}^-_1 + \omega_{1t} \hat{c}^\dagger \hat{X}^-_1 )\rho_s +(\nu_{2t} \hat{X}^+_2\hat{c} -\omega_{2t}\hat{c}^\dagger \hat{X}^-_2)\rho_s $$ In above equation the 2nd term on R.H.S can be written as $(\nu_{1t}+ \omega_{1t} ) \hat{c}^\dagger \hat{X}^-_1 \rho_s$ . Now $(\nu_{1t}+ \omega_{1t} )$ can be replaced by another noise $\alpha_{t}$ whose auto and cross-correlation can be found. $\endgroup$ Dec 27, 2018 at 8:37

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Thank you for your clarification. The best I can propose is to write the operators in terms of a quadratic form $$\left[\begin{array}{ccc} c^{\dagger} & X_{1}^{+} & X_{2}^{+}\end{array}\right]Q\left[\begin{array}{c} c\\ X_{1}^{-}\\ X_{2}^{-} \end{array}\right]\rho_{s}$$ , where $Q$ is a $3x3$ matrix that contains the $c$-number noise terms. as you see, there are 9 entries, one for each possible operator combination. in your example case in the comments, the form is $$\left[\begin{array}{ccc} c^{\dagger} & X_{1}^{+} & X_{2}^{+}\end{array}\right]\left[\begin{array}{ccc} 0 & \nu_{1}+\omega_1 & -\omega_{2}\\ 0 & 0 & 0\\ \nu_2 & 0 & 0 \end{array}\right]\left[\begin{array}{c} c\\ X_{1}^{-}\\ X_{2}^{-} \end{array}\right]\rho_{s}$$ but in the question the form is $$\left[\begin{array}{ccc} c^{\dagger} & X_{1}^{+} & X_{2}^{+}\end{array}\right]\left[\begin{array}{ccc} 0 & \nu_{1} & -\omega_{2}\\ \omega_{1} & 0 & 0\\ \nu_{2} & 0 & 0 \end{array}\right]\left[\begin{array}{c} c\\ X_{1}^{-}\\ X_{2}^{-} \end{array}\right]\rho_{s}$$ and there are 4 noise terms. You could consider some kind of similarity transformation on the $Q$ but it seems to me that the transformation would have to be noise dependent.

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  • $\begingroup$ Thanks for your response, will this similarity transformation decrease the number of noises ? $\endgroup$ Dec 28, 2018 at 8:01

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